Problem: Solve for $n$, $ -\dfrac{3}{4n - 3} = \dfrac{n + 6}{4n - 3} + \dfrac{6}{20n - 15} $
Solution: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $4n - 3$ $4n - 3$ and $20n - 15$ The common denominator is $20n - 15$ To get $20n - 15$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ -\dfrac{3}{4n - 3} \times \dfrac{5}{5} = -\dfrac{15}{20n - 15} $ To get $20n - 15$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ \dfrac{n + 6}{4n - 3} \times \dfrac{5}{5} = \dfrac{5n + 30}{20n - 15} $ The denominator of the third term is already $20n - 15$ , so we don't need to change it. This give us: $ -\dfrac{15}{20n - 15} = \dfrac{5n + 30}{20n - 15} + \dfrac{6}{20n - 15} $ If we multiply both sides of the equation by $20n - 15$ , we get: $ -15 = 5n + 30 + 6$ $ -15 = 5n + 36$ $ -51 = 5n $ $ n = -\dfrac{51}{5}$